statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYS7076
物理代写|量子场论代写Quantum field theory代考|Covariant equation of motion
The correct relativistic version of Newton’s equation of motion can be formulated in terms of the 4-momentum and the proper time as follows, $$ \frac{d p^\mu}{d \tau}=K^\mu . $$ The quantity $K^\mu$ is a 4-vector, denoted as 4-force. Since all terms transform with the same matrix $\Lambda$, a Lorentz transformation into another reference frame, with $p^\mu \rightarrow p^{\prime \mu}, K^\mu \rightarrow K^{\prime \mu}$, yields the transformed equation of motion $$ \frac{d p^{\prime \mu}}{d \tau}=K^{\prime \mu} . $$ The individual components have changed, but the equation has the same form in different inertial frames; the equation is invariant in form, or covariant. In this context, covariant means that an equation can be written so that both sides have the same, well-defined, transformation properties under Lorentz transformations. This is an example for describing a physical law with the help of vector or tensor components, making Lorentz symmetry manifest. The 3-dimensional part of Eq. (1.44) can be cast into the form $$ \frac{d p^k}{d t}=K^k \sqrt{1-\frac{\vec{v}^2}{c^2}} \equiv F^k, $$ involving the 3 -dimensional force $\vec{F}$; an example will be given below. The physical meaning of the 0-component of the 4 -force follows from the indentity $$ p_\mu \frac{d p^\mu}{d \tau}=\frac{1}{2} \frac{d p^2}{d \tau}=0 \quad\left(\text { since } p^2=m^2 c^2\right) $$ together with $$ p_\mu \frac{d p^\mu}{d \tau}=p_\mu K^\mu=p_0 K^0-\vec{p} \vec{K}=0 $$ to become $$ p_0 K^0=\vec{p} \vec{K} . $$
物理代写|量子场论代写Quantum field theory代考|Free particle
Consider the free motion of a particle with mass $m$, described by the coordinates $x_i$ and the velocities $\dot{x}_i=v_i$. The corresponding Lagrangian is given by $$ L=-m \sqrt{1-\vec{v}^2} . $$ The equations of motions follow as the Euler-Lagrange equations, $$ \frac{d}{d t}\left(\frac{\partial L}{\partial v_i}\right)-\frac{\partial L}{\partial x_i}=0 $$ yielding $$ \frac{d}{d t}\left(\frac{m v_i}{\sqrt{1-\vec{v}^2}}\right)=0, \quad \frac{d \vec{p}}{d t}=0 . $$ They coincide with Eq. (1.46) for $\vec{F}=0$ and confirm Eq. (1.51) as the correct relativistic version for the Lagrangian of a free particle.
Hamiltonian. The Hamiltonian $H$ is derived from the Lagrangian $L$ by means of a Legrendre transformation, $$ H=\sum_i p_i v_i-L $$ with the canonical momenta $$ p_i=\frac{\partial L}{\partial v_i}, $$ where the velocities $v_i$ have to be substituted by the momenta $p_i$.
statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYS3101
物理代写|量子场论代写Quantum field theory代考|Lorentz Transformations
A Lorentz transformation from an inertial frame $K$ into another inertial frame $K^{\prime}$ is described by a linear transformation of the components $x^\mu$ in $K$ to $x^{\prime \mu}$ in $K^{\prime}$ with the help of a matrix $\Lambda=\left(\Lambda_\nu^\mu\right)$, $$ x^{\prime \mu}=\Lambda_\nu^\mu x^\nu \text {. } $$ The invariance of light propagation (1.1) is equivalent to the invariance of the scalar product under Lorentz transformations, which can be fomulated as a condition $$ g_{\rho \sigma} x^\rho x^\sigma=g_{\mu \nu} x^{\prime \mu} x^{\prime \nu}=g_{\mu \nu} \Lambda_\rho^\mu \Lambda_\sigma^\nu x^\rho x^\sigma $$ to be fulfilled for all $x^\mu$. Hence, Eq. (1.11) is a condition for the entries of $\Lambda$, $$ g_{\mu \nu} \Lambda_\rho^\mu \Lambda_\sigma^\nu=g_{\rho \sigma}, $$ or expressed in a compact way using the matrix $g=\left(g_{\mu \nu}\right)$ as follows, $$ \Lambda^T g \Lambda=g . $$ Applying the rules for determinants, $$ \operatorname{det}\left(\Lambda^T g \Lambda\right)=\operatorname{det}\left(\Lambda^T\right) \operatorname{det}(g) \operatorname{det}(\Lambda)=\operatorname{det}(g), $$ it immediately follows for the determinant of a Lorentz transformation that $$ \operatorname{det}(\Lambda)=\pm 1 . $$ As a consequence, the 4-dimensional volume element is invariant under Lorentz transformations, $$ \mathrm{d}^4 x=\mathrm{d} x^0 \mathrm{~d}^3 x \quad \rightarrow \quad \mathrm{d}^4 x^{\prime}=|\operatorname{det}(\Lambda)| \mathrm{d}^4 x=\mathrm{d}^4 x, $$ because $\Lambda$ is the Jacobian matrix of the transformation of variables $x^\mu \rightarrow x^{\prime \mu}$
物理代写|量子场论代写Quantum field theory代考|Mechanics
The motion of a particle with mass $m$ along a given trajectory is described by a time-dependent 3-dimensional vector $\vec{x}(t)$, with components $x^k(t)$, $k=1,2,3$. The velocity is as usual defined as the derivative $$ \vec{v}=\frac{d \vec{x}}{d t}=\left(v^1, v^2, v^3\right), \quad v^k=\frac{d x^k}{d t} . $$ The 4-dimensional trajectory is described by a curve in Minkowski space, $$ t \rightarrow x^\mu(t) \text { with } x^0(t)=c t, $$ denoted as the world line of the particle. For neighbouring points $x^\mu$ und $x^\mu+d x^\mu$ along the world line the difference is given by $$ d x^\mu=\frac{d x^\mu}{d t} d t $$ yielding the invariant 4-dimensional square, the line element of the world line, $$ \begin{aligned} d s^2 &=g_{\mu \nu} d x^\mu d x^\nu=g_{\mu \nu} \frac{d x^\mu}{d t} \frac{d x^\nu}{d t} d t^2 \ &=\left[\left(\frac{d x^0}{d t}\right)^2-\vec{v}^2\right] d t^2=c^2 d t^2\left[1-\frac{\vec{v}^2}{c^2}\right] \equiv c^2 d \tau^2 \end{aligned} $$
By this relation the invariant interval of the proper time is defined, $$ d \tau=d t \sqrt{1-\frac{\vec{v}^2}{c^2}} . $$ It is displayed by a co-moving clock.
statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYSICS3544
物理代写|量子场论代写Quantum field theory代考|Special Relativity
It is a assumed that the reader is already familiar with special relativity, e.g. at the level of a first course on electrodynamics. Thus, this chapter serves essentially as a summary of the relevant theoretical ingredients, to introduce notations and conventions, and to provide the classical basis for advancing to quantum theory in the subseqent chapters.
The Special Theory of Relativity is based on the following two fundamental principles. (1) Principle of relativity The laws of nature are in all inertial frames of the same form. (2) Invariance of light propagation The propagation of light is independent of the inertial frame. For a pointlike source at a space point $\vec{x}0$ and time $t_0$ in a given frame $K$ the outgoing lightfronts are spherical surfaces in all inertial frames, and propagate with $c$, the constant velocity of light, $$ \underbrace{c^2\left(t-t_0\right)^2-\left(\vec{x}-\vec{x}_0\right)^2}{\text {inertial frame } K}=0=\underbrace{c^2\left(t^{\prime}-t_0^{\prime}\right)^2-\left(\vec{x}^{\prime}-\vec{x}0^{\prime}\right)^2}{\text {inertial frame } K^{\prime}} . $$ Principle (2) is not valid for Galilei transformations between different frames. Thus, they have to be replaced by Lorentz transformations.
物理代写|量子场论代写Quantum field theory代考|Notations and Conventions
In a given inertial frame a space-time event is determined by a set of coordinates specifying the point $\vec{x}$ in space and $t$ in time. This information can be summarized by a four-component quantity $\left(x^\mu\right)$ according to $$ \left(x^\mu\right)=\left(x^0, x^1, x^2, x^3\right) \equiv\left(x^0, \vec{x}\right) $$ with $$ x^0=c t, \quad \vec{x}=\left(x^1, x^2, x^3\right) . $$ The $\left(x^\mu\right)$ form a 4-dimensional linear space, with elements denoted as 4-vectors and labeled by the symbol $x \equiv\left(x^\mu\right)$ as a short-hand notation. The quantities $x^\mu$ are called contravariant components. Convention: greek indices $\mu, \nu, \ldots=0,1,2,3$ latin indices $k, l, \ldots=1,2,3$. Metric. For the space of 4-vectors a scalar product is defined as a symmetric bilinear form as follows, $$ (x, y) \rightarrow x \cdot y=x^0 y^0-\vec{x} \cdot \vec{y}=g_{\mu \nu} x^\mu y^\nu $$ with the metric tensor $$ \left(g_{\mu \nu}\right)=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & -1 \end{array}\right) $$ The square of a 4-vector is thus given by $$ x^2=x \cdot x=g_{\mu \nu} x^\mu x^\nu=\left(x^0\right)^2-\vec{x}^2 . $$ The 4-dimensional space of the $\left(x^\mu\right)$ with this metric is denoted as Minkowski space, the metric (1.4) as Minkowski metric.
statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYS8302
物理代写|量子场论代写Quantum field theory代考|Thermal Correlation Functions
The energies of excited states are encoded in the thermal correlation functions. These functions are expectation values of products of the position operator $$ \hat{q}{\mathrm{E}}(\tau)=\mathrm{e}^{\tau \hat{H} / \hbar} \hat{q} \mathrm{e}^{-\tau \hat{H} / \hbar}, \quad \hat{q}{\mathrm{E}}(0)=\hat{q}(0), $$ at different imaginary times in the canonical ensemble, $$ \left\langle\hat{q}{\mathrm{E}}\left(\tau{1}\right) \cdots \hat{q}{\mathrm{E}}\left(\tau{n}\right)\right\rangle_{\beta} \equiv \frac{1}{Z(\beta)} \operatorname{tr}\left(\mathrm{e}^{-\beta \hat{H}} \hat{q}{\mathrm{E}}\left(\tau{1}\right) \cdots \hat{q}{\mathrm{E}}\left(\tau{n}\right)\right) $$ The normalizing function $Z(\beta)$ is the partition function (2.56). From the thermal two-point function $$ \begin{aligned} \left\langle\hat{q}{\mathrm{E}}\left(\tau{1}\right) \hat{q}{\mathrm{E}}\left(\tau{2}\right)\right\rangle_{\beta} &=\frac{1}{Z(\beta)} \operatorname{tr}\left(\mathrm{e}^{-\beta \hat{H}} \hat{q}{\mathrm{E}}\left(\tau{1}\right) \hat{q}{\mathrm{E}}\left(\tau{2}\right)\right) \ &=\frac{1}{Z(\beta)} \operatorname{tr}\left(\mathrm{e}^{-\left(\beta-\tau_{1}\right) \hat{H}} \hat{q} \mathrm{e}^{-\left(\tau_{1}-\tau_{2}\right) \hat{H}} \hat{q} \mathrm{e}^{-\tau_{2} \hat{H}}\right) \end{aligned} $$ we can extract the energy gap between the ground state and the first excited state. For this purpose we use orthonormal energy eigenstates $|n\rangle$ to calculate the trace and in addition insert the resolution of the identity operator $\mathbb{1}=\sum|m\rangle\langle m|$. This yields $$ \langle\ldots\rangle_{\beta}=\frac{1}{Z(\beta)} \sum_{n, m} \mathrm{e}^{-\left(\beta-\tau_{1}+\tau_{2}\right) E_{n}} \mathrm{e}^{-\left(\tau_{1}-\tau_{2}\right) E_{\mathrm{m}}}\langle n|\hat{q}| m\rangle\langle m|\hat{q}| n\rangle $$ Note that in the sum over $n$ the contributions from the excited states are exponentially suppressed at low temperatures $\beta \rightarrow \infty$, implying that the thermal two-point function converges to the Schwinger function in this limit: $$ \left\langle\hat{q}{\mathrm{E}}\left(\tau{1}\right) \hat{q}{\mathrm{E}}\left(\tau{2}\right)\right\rangle_{\beta} \stackrel{\beta \rightarrow \infty}{\longrightarrow} \sum_{m>0} \mathrm{e}^{-\left(\tau_{1}-\tau_{2}\right)\left(E_{m}-E_{0}\right)}|\langle 0|\hat{q}| m\rangle|^{2}=\left\langle 0\left|\hat{q}{\mathrm{E}}\left(\tau{1}\right) \hat{q}{\mathrm{E}}\left(\tau{2}\right)\right| 0\right\rangle $$
物理代写|量子场论代写Quantum field theory代考|The Harmonic Oscillator
We wish to study the path integral for the Euclidean oscillator with discretized time. The oscillator is one of the few systems for which the path integral can be calculated explicitly. For more such system, the reader may consult the text [19]. But the results for the oscillator are particularly instructive with regard to lattice field theories considered later in this book. So let us discretize the Euclidean time interval $[0, \tau]$ with $n$ sampling points separated by a lattice constant $\varepsilon=\tau / n$. For the Lagrangian $$ L=\frac{m}{2} \dot{q}^{2}+\mu q^{2} $$ the discretized path integral over periodic paths reads $$ \begin{aligned} Z &=\int \mathrm{d} q_{1} \cdots \mathrm{d} q_{n}\left(\frac{m}{2 \pi \varepsilon}\right)^{n / 2} \exp \left{-\varepsilon \sum_{j=0}^{n-1}\left(\frac{m}{2}\left(\frac{q_{j+1}-q_{j}}{\varepsilon}\right)^{2}+\mu q_{j}^{2}\right)\right} \ &=\left(\frac{m}{2 \pi \varepsilon}\right)^{n / 2} \int \mathrm{d} q_{1} \cdots \mathrm{d} q_{n} \exp \left(-\frac{1}{2}(\boldsymbol{q}, \mathrm{A} q)\right) \end{aligned} $$ where we assumed $q_{0}=q_{n}$ and introduced the symmetric matrix $$ \mathrm{A}=\frac{m}{\varepsilon}\left(\begin{array}{cccccc} \alpha & -1 & 0 & \cdots & 0 & -1 \ -1 & \alpha & -1 & \cdots & 0 & 0 \ & & \ddots & & & \ & & & \ddots & & \ 0 & 0 & \cdots & -1 & \alpha & -1 \ -1 & 0 & \cdots & 0 & -1 & \alpha \end{array}\right), \quad \alpha=2\left(1+\frac{\mu}{m} \varepsilon^{2}\right) $$ This is a Toeplitz matrix in which each descending diagonal from left to right is constant. This property results from the invariance of the action under lattice translations. For the explicit calculation of $Z$, we consider the generating function $$ \begin{aligned} Z[j] &=\left(\frac{m}{2 \pi \varepsilon}\right)^{n / 2} \int \mathrm{d}^{n} q \exp \left{-\frac{1}{2}(\boldsymbol{q}, \mathrm{A} q)+(\boldsymbol{j}, \boldsymbol{q})\right} \ &=\frac{(m / \varepsilon)^{n / 2}}{\sqrt{\operatorname{det} \mathrm{A}}} \exp \left{\frac{1}{2}\left(j, \mathrm{~A}^{-1} \boldsymbol{j}\right)\right} \end{aligned} $$
物理代写|量子场论代写Quantum field theory代考|Problems
2.1 (Gaussian Integral) Show that $$ \int \mathrm{d} z_{1} \mathrm{~d} \bar{z}{1} \ldots \mathrm{d} z{n} \mathrm{~d} \bar{z}{n} \exp \left(-\sum{i j} \bar{z}{i} A{i j} z_{j}\right)=\pi^{n}(\operatorname{det} \mathrm{A})^{-1} $$ with A being a positive Hermitian $n \times n$ matrix and $z_{i}$ complex integration variables. 2.2 (Harmonic Oscillator) In (2.43) we quoted the result for the kernel $K_{\omega}\left(\tau, q^{\prime}, q\right)$ of the $d$-dimensional harmonic oscillator with Hamiltonian $$ \hat{H}=\frac{1}{2 m} \hat{p}^{2}+\frac{m \omega^{2}}{2} \hat{q}^{2} $$ at imaginary time $\tau$. Derive this formula. Hint: Express the kernel in terms of the eigenfunctions of $\hat{H}$, which for $\hbar=m=$ $\omega=1$ are given by $$ \exp \left(-\xi^{2}-\eta^{2}\right) \sum_{n=0}^{\infty} \frac{\alpha^{n}}{2^{n} n !} H_{n}(\xi) H_{n}(\eta)=\frac{1}{\sqrt{1-\alpha^{2}}} \exp \left(\frac{-\left(\xi^{2}+\eta^{2}-2 \xi \eta \alpha\right)}{1-\alpha^{2}}\right) $$ The functions $H_{n}$ denote the Hermite polynomials. Comment: This result also follows from the direct evaluation of the path integral. 2.3 (Free Particle on a Circle) A free particle moves on an interval and obeys periodic boundary conditions. Compute the time evolution kernel $K\left(t_{b}-t_{a}, q_{b}, q_{a}\right)=$ $\left\langle q_{b}, t_{b} \mid q_{a}, t_{a}\right\rangle$. Use the familiar formula for the kernel of the free particle (2.26) and enforce the periodic boundary conditions by a suitable sum over the evolution kernel for the particle on $\mathbb{R}$.
statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYS3101
物理代写|量子场论代写Quantum field theory代考|Quantum Mechanics in Imaginary Time
The unitary time evolution operator has the spectral representation $$ \hat{K}(t)=\mathrm{e}^{-\mathrm{i} \hat{H} t}=\int \mathrm{e}^{-\mathrm{i} E t} \mathrm{~d} \hat{P}{\mathrm{E}}, $$ where $\hat{P}{\mathrm{E}}$ is the spectral family of the Hamiltonian. If $\hat{H}$ has discrete spectrum, then $\hat{P}{\mathrm{E}}$ is the orthogonal projector onto the subspace of $\mathscr{H}$ spanned by all eigenfunctions with energies less than $E$. In the following we assume that the Hamiltonian operator is bounded from below. Then we can subtract its ground state energy to obtain a non-negative $\hat{H}$ for which the integration limits in (2.35) are 0 and $\infty$. With the substitution $t \rightarrow t-\mathrm{i} \tau$, we obtain $$ \mathrm{e}^{-(\tau+\mathrm{i} t) \hat{H}}=\int{0}^{\infty} \mathrm{e}^{-E(\tau+\mathrm{i} t)} \mathrm{d} \hat{P}_{\mathrm{E}} $$ This defines a holomorphic semigroup in the lower complex half-plane $$ {z=t-\mathrm{i} \tau \in \mathbb{C}, \tau \geq 0} $$ If the operator $(2.36)$ is known on the negative imaginary axis $(t=0, \tau \geq 0)$, one can perform an analytic continuation to the real axis $(t, \tau=0)$. The analytic continuation to complex time $t \rightarrow-\mathrm{i} \tau$ corresponds to a transition from the Minkowski metric $\mathrm{d} s^{2}=d t^{2}-\mathrm{d} x^{2}-\mathrm{d} y^{2}-\mathrm{d} z^{2}$ to a metric with Euclidean signature. Hence a theory with imaginary time is called Euclidean theory.
The time evolution operator $\hat{K}(t)$ exists for real time and defines a oneparametric unitary group. It fulfills the Schrödinger equation $$ \mathrm{i} \frac{\mathrm{d}}{\mathrm{d} t} \hat{K}(t)=\hat{H} \hat{K}(t) $$ with a complex and oscillating kernel $K\left(t, q^{\prime}, q\right)=\left\langle q^{\prime}|\hat{K}(t)| q\right\rangle$. For imaginary time we have a Hermitian (and not unitary) evolution operator $$ \hat{K}(\tau)=\mathrm{e}^{-\tau \hat{H}} $$ with positive spectrum. The $\hat{K}(\tau)$ exist for positive $\tau$ and form a semi-group only. For almost all initial data, evolution back into the “imaginary past” is impossible. The evolution operator for imaginary time satisfies the heat equation $$ \frac{\mathrm{d}}{\mathrm{d} \tau} \hat{K}(\tau)=-\hat{H} \hat{K}(\tau) $$ instead of the Schrödinger equation and has kernel $$ K\left(\tau, q^{\prime}, q\right)=\left\langle q^{\prime}\left|\mathrm{e}^{-\tau \hat{H}}\right| q\right\rangle, \quad K\left(0, q^{\prime}, q\right)=\delta\left(q^{\prime}, q\right) $$ This kernel is real ${ }^{1}$ for a real Hamiltonian. Furthermore it is strictly positive:
物理代写|量子场论代写Quantum field theory代考|Imaginary Time Path Integral
To formulate the path integral for imaginary time, we employ the product formula $(2.28)$, which follows from the product formula (2.27) through the substitution of it by $\tau$. For such systems the analog of $(2.31)$ for Euclidean time $\tau$ is obtained by the substitution of $i \varepsilon$ by $\varepsilon$. Thus we tind $$ \begin{aligned} K\left(\tau, q^{\prime}, q\right) &=\left\langle\hat{q}^{\prime}\left|\mathrm{e}^{-\tau \hat{H} / \hbar}\right| \hat{q}\right\rangle \ &=\lim {n \rightarrow \infty} \int \mathrm{d} q{1} \cdots \mathrm{d} q_{n-1}\left(\frac{m}{2 \pi \hbar \varepsilon}\right)^{n / 2} \mathrm{e}^{-S_{\mathrm{E}}\left(q_{0}, q_{1}, \ldots, q_{n}\right) / \hbar} \ S_{\mathrm{E}}(\ldots) &=\varepsilon \sum_{j=0}^{n-1}\left{\frac{m}{2}\left(\frac{q_{j+1}-q_{j}}{\varepsilon}\right)^{2}+V\left(q_{j}\right)\right} \end{aligned} $$ where $q_{0}=q$ and $q_{n}=q^{\prime}$. The multidimensional integral represents the sum over all broken-line paths from $q$ to $q^{\prime}$. Interpreting $S_{\mathrm{E}}$ as Hamiltonian of a classical lattice model and $\hbar$ as temperature, it is (up to the fixed endpoints) the partition function of a one-dimensional lattice model on a lattice with $n+1$ sites. The realvalued variable $q_{j}$ defined on site $j$ enters the action $S_{\mathrm{E}}$ which contains interactions between the variables $q_{j}$ and $q_{j+1}$ at neighboring sites. The values of the lattice field $$ {0,1, \ldots, n-1, n} \rightarrow\left{q_{0}, q_{1}, \ldots, q_{n-1}, q_{n}\right} $$ are prescribed at the endpoints $q_{0}=q$ and $q_{n}=q^{\prime}$. Note that the classical limit $\hbar \rightarrow 0$ corresponds to the low-temperature limit of the lattice system.
The multidimensional integral (2.52) corresponds to the summation over all path on the time lattice. What happens to the finite-dimensional integral when we take the continuum limit $n \rightarrow \infty$ ? Then we obtain the Euclidean path integral representation for the positive kernel $$ K\left(\tau, q^{\prime}, q\right)=\left\langle q^{\prime}\left|\mathrm{e}^{-\tau \hat{H} / h}\right| q\right\rangle=C \int_{q(0)=q}^{q(\tau)=q^{\prime}} \mathscr{D} q \mathrm{e}^{-S_{\mathrm{E}}[q] / h} $$ The integrand contains the Euclidean action $$ S_{\mathrm{E}}[q]=\int_{0}^{\tau} d \sigma\left{\frac{m}{2} \dot{q}^{2}+V(q(\sigma))\right} $$ which for many physical systems is bounded from below.
物理代写|量子场论代写Quantum field theory代考|Path Integral in Quantum Statistics
The Euclidean path integral formulation immediately leads to an interesting connection between quantum statistical mechanics and classical statistical physics. Indeed, if we set $\tau / \hbar \equiv \beta$ and integrate over $q=q^{\prime}$ in (2.53), then we end up with the path integral representation for the canonical partition function of a quantum system with Hamiltonian $\hat{H}$ at inverse temperature $\beta=1 / k_{B} T$. More precisely, setting $q=q^{\prime}$ and $\tau=\hbar \beta$ in the left-hand side of this formula, then the integral over $q$ yields the trace of $\exp (-\beta \hat{H})$, which is just the canonical partition function, $$ \int \mathrm{d} q K(\hbar \beta, q, q)=\operatorname{tr} \mathrm{e}^{-\beta \hat{H}}=Z(\beta)=\sum \mathrm{e}^{-\beta E_{n}} \quad \text { with } \quad \beta=\frac{1}{k_{B} T} $$ Setting $q=q^{\prime}$ in the Euclidean path integral in (2.53) means that we integrate over paths beginning and ending at $q$ during the imaginary time interval $[0, \hbar \beta]$. The final integral over $q$ leads to the path integral over all periodic paths with period $\hbar \beta$ $$ Z(\beta)=C \oint \mathscr{D} q \mathrm{e}^{-S_{\mathrm{E}}[q] / \hbar}, \quad q(\hbar \beta)=q(0) $$ For example, the kernell of the harmonic oscillator in $(2.43)$ on the diagonal is $$ K_{\omega}(\beta, q, q)=\sqrt{\frac{m \omega}{2 \pi \sinh (\omega \beta)}} \exp \left{-m \omega \tanh (\omega \beta / 2) q^{2}\right} $$ where we used units with $\hbar=1$. The integral over $q$ yields the partition function $$ \begin{aligned} Z(\beta) &=\sqrt{\frac{m \omega}{2 \pi \sinh (\omega \beta)}} \int \mathrm{d} q \exp \left{-m \omega \tanh (\omega \beta / 2) q^{2}\right} \ &=\frac{1}{2 \sinh (\omega \beta / 2)}=\frac{\mathrm{e}^{-\omega \beta / 2}}{1-\mathrm{e}^{-\omega \beta}}=\mathrm{e}^{-\omega \beta / 2} \sum_{n=0}^{\infty} \mathrm{e}^{-n \omega \beta} \end{aligned} $$ where we used $\sinh x=2 \sinh x / 2 \cosh x / 2$. A comparison with the spectral sum over all energies in (2.55) yields the energies of the oscillator with (angular) frequency $\omega$, $$ E_{n}=\omega\left(n+\frac{1}{2}\right), \quad n=0,1,2, \ldots $$ For large values of $\omega \beta$, i.e., for very low temperature, the spectral sum is dominated by the contribution of the ground state energy. Thus for cold systems, the free energy converges to the ground state energy $$ F(\beta) \equiv-\frac{1}{\beta} \log Z(\beta) \stackrel{\omega \beta \rightarrow \infty}{\longrightarrow} E_{0} $$ One often is interested in the energies and wave functions of excited states. We now discuss an elegant method to extract this information from the path integral.
S_{\mathrm{E}}[q]=\int_{0}^{\tau} d \sigma\left{\frac{m}{2} \dot{q}^{2}+V(q( \sigma))\对}S_{\mathrm{E}}[q]=\int_{0}^{\tau} d \sigma\left{\frac{m}{2} \dot{q}^{2}+V(q( \sigma))\对} 对于许多物理系统来说，它是从下面限定的。
物理代写|量子场论代写Quantum field theory代考|Path Integral in Quantum Statistics
statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYSICS 3544
物理代写|量子场论代写Quantum field theory代考|Path Integrals in Quantum and Statistical Mechanics
Already back in 1933 , Dirac asked himself whether the classical Lagrangian and action are as significant in quantum mechanics as they are in classical mechanics $[1,2]$. He observed that for simple systems, the probability amplitude $$ K\left(t, q^{\prime}, q\right)=\left\langle q^{\prime}\left|\mathrm{e}^{-\mathrm{i} \hat{A} t / h}\right| q\right\rangle $$ for the propagation from a point with coordinate $q$ to another point with coordinate $q^{\prime}$ in time $t$ is given by $$ K\left(t, q^{\prime}, q\right) \propto \mathrm{e}^{\mathrm{i} S\left[q_{\mathrm{cl}}\right] / h} $$ where $q_{\mathrm{cl}}$ denotes the classical trajectory from $q$ to $q^{\prime}$. In the exponent the action of this trajectory enters as a multiple of Planck’s reduced constant $h$. For a free particle with Lagrangian $$ L_{0}=\frac{m}{2} \dot{q}^{2} $$ the formula $(2.2)$ is verified easily: A free particle moves with constant velocity $\left(q^{\prime}-q\right) / t$ from $q$ to $q^{\prime}$ and the action of the classical trajectory is $$ S\left[q_{\mathrm{cl}}\right]=\int_{0}^{t} \mathrm{~d} s L_{0}\left[q_{\mathrm{cl}}(s)\right]=\frac{m}{2 t}\left(q^{\prime}-q\right)^{2} $$ The factor of proportionality in $(2.2)$ is then uniquely fixed by the condition $\mathrm{e}^{-\mathrm{i} \hat{H} t / \hbar} \longrightarrow 1$ for $t \rightarrow 0$ which in position space reads $$ \lim {t \rightarrow 0} K\left(t, q^{\prime}, q\right)=\delta\left(q^{\prime}, q\right) $$ Alternatively, it is fixed by the property $\mathrm{e}^{-\mathrm{i} \hat{H} t / h} \mathrm{e}^{-\mathrm{i} \hat{H} s / h}=\mathrm{e}^{-\mathrm{i} \hat{H}(t+s) / h}$ that takes the form $$ \int \mathrm{d} u K\left(t, q^{\prime}, u\right) K(s, u, q)=K\left(t+s, q^{\prime}, q\right) $$ in position space. Thus, the correct free particle propagator on a line is given by $$ K{0}\left(t, q^{\prime} \cdot q\right)=\left(\frac{m}{2 \pi \mathrm{i} \hbar t}\right)^{1 / 2} \mathrm{c}^{\mathrm{i} m\left(q^{\prime}-q\right)^{2} / 2 h t} $$ Similar results hold for the harmonic oscillator or systems for which $\langle\hat{q}(t)\rangle$ fulfills the classical equation of motion. For such systems $\left\langle V^{\prime}(\hat{q})\right\rangle=V^{\prime}(\langle\hat{q}\rangle)$ holds true. However, for general systems, the simple formula (2.2) must be extended, and it was Feynman who discovered this extension back in 1948. He realized that all paths from $q$ to $q^{\prime}$ (and not only the classical path) contribute to the propagator. This means that in quantum mechanics a particle can potentially move on any path $q(s)$ from the initial to the final destination, $$ q(0)=q \quad \text { and } \quad q(t)=q^{\prime} $$
物理代写|量子场论代写Quantum field theory代考|Recalling Quantum Mechanics
There are two well-established ways to quantize a classical system: canonical quantization and path integral quantization. For completeness and later use, we recall the main steps of canonical quantization both in Schrödinger’s wave mechanics and Heisenberg’s matrix mechanics.
A classical system is described by its coordinates $\left{q^{i}\right}$ and momenta $\left{p_{i}\right}$ on phase space $\Gamma$. An observable $O$ is a real-valued function on $\Gamma$. Examples are the coordinates on phase space and the energy $H(q, p)$. We assume that phase space comes along with a symplectic structure and has local coordinates with Poisson brackets $$ \left{q^{i}, p_{j}\right}=\delta_{j}^{i} $$ The brackets are extended to observables on through antisymmetry and the derivation rule ${O P, Q}=O{P, Q}+{O, Q} P$. The evolution in time of an observable is determined by $$ \dot{O}={O, H}, \quad \text { e.g. } \quad \dot{q}^{i}=\left{q^{i}, H\right} \quad \text { and } \quad \dot{p}{i}=\left{p{i}, H\right} $$ In the canonical quantization, functions on phase space are mapped to operators, and the Poisson brackets of two functions become commutators of the associated operators: $$ O(q, p) \rightarrow \hat{O}(\hat{q}, \hat{p}) \quad \text { and } \quad{O, P} \longrightarrow \frac{1}{\mathrm{i} \hbar}[\hat{O}, \hat{P}] $$
The time evolution of an (not explicitly time-dependent) observable is determined by Heisenberg’s equation $$ \frac{\mathrm{d} \hat{O}}{\mathrm{~d} t}=\frac{\mathrm{i}}{\hbar}[\hat{H}, \hat{O}] $$ In particular the phase space coordinates $\left(q^{l}, p_{i}\right)$ become operators with commutation relations $\left[\hat{q}^{i}, \hat{p}{j}\right]=\mathrm{i} \hbar \delta{j}^{i}$, and their time evolution is determined by $$ \frac{\mathrm{d} \hat{q}^{i}}{\mathrm{~d} t}=\frac{\mathrm{i}}{\hbar}\left[\hat{H}, \hat{q}^{i}\right] \quad \text { and } \quad \frac{\mathrm{d} \hat{p}{i}}{\mathrm{~d} t}=\frac{\mathrm{i}}{\hbar}\left[\hat{H}, \hat{p}{i}\right] $$ For a system of non-relativistic and spinless particles, the Hamiltonian reads $$ \hat{H}=\hat{H}{0}+\hat{V} \quad \text { with } \quad \hat{H}{0}=\frac{1}{2 m} \sum \hat{p}{i}^{2} $$ and one arrives at Heisenberg’s equations of motion $$ \frac{\mathrm{d} \hat{q}^{i}}{\mathrm{~d} t}=\frac{\hat{p}{i}}{m} \quad \text { and } \quad \frac{\mathrm{d} \hat{p}{i}}{\mathrm{~d} t}=-\hat{V}{, i} $$ Observables are represented by Hermitian operators on a Hilbert space $\mathscr{H}$, whose elements characterize the states of the system: $$ \hat{O}(\hat{q}, \hat{p}): \mathcal{H} \longrightarrow \mathcal{H} $$ Consider a particle confined to an endless wire. Its Hilbert space is $\mathcal{H}=L_{2}(\mathbb{R})$, and its position and momentum operator are represented in position space as $$ (\hat{q} \psi)(q)=q \psi(q) \quad \text { and } \quad(\hat{p} \psi)(q)=\frac{\hbar}{i} \partial_{q} \psi(q) $$
物理代写|量子场论代写Quantum field theory代考|Feynman–Kac Formula
We shall derive Feynman’s path integral representation for the unitary time evolution operator $\exp (-\mathrm{i} \hat{H} t)$ as well as Kac’s path integral representation for the positive operator $\exp (-\hat{H} \tau)$. Thereby we shall utilize the product formula of Trotter. In case of matrices, this formula was already verified by Lie and has the form: Theorem 2.1 (Lie’s Theorem) For two matrices $\mathrm{A}$ and $\mathrm{B}$ $$ \mathrm{e}^{\mathrm{A}+\mathrm{B}}=\lim {n \rightarrow \infty}\left(\mathrm{e}^{\mathrm{A} / n} \mathrm{e}^{\mathrm{B} / n}\right)^{n} $$ To prove this theorem, we define for each $n$ the two matrices $\mathrm{S}{n}:=\exp (\mathrm{A} / n+\mathrm{B} / n)$ and $\mathrm{T}{n}:=\exp (\mathrm{A} / n) \exp (\mathrm{B} / n)$ and telescope the difference of their $n$ ‘th powers, $$ \mathrm{S}{n}^{n}-\mathrm{T}{n}^{n}=\mathrm{S}{n}^{n-1}\left(\mathrm{~S}{n}-\mathrm{T}{n}\right)+\mathrm{S}{n}^{n-2}\left(\mathrm{~S}{n}-\mathrm{T}{n}\right) \mathrm{T}{n}+\cdots+\left(\mathrm{S}{n}-\mathrm{T}{n}\right) \mathrm{T}{n}^{n-1} $$ Now we choose any (sub-multiplicative) matrix norm, for example, the Frobenius norm. The triangle inequality together with $|X Y| \leq|X \mid| Y |$ imply the inequality $|\exp (X)| \leq \exp (|X|)$ such that $$ \left|\mathrm{S}{n}\right|,\left|\mathrm{T}{n}\right| \leq a^{1 / n} \quad \text { with } \quad a=\mathrm{e}^{|\mathrm{A}|+|\mathrm{B}|} $$ Thus we conclude $$ \left|\mathrm{S}{n}^{n}-\mathrm{T}{n}^{n}\right| \equiv\left|\mathrm{e}^{\mathrm{A}+B}-\left(\mathrm{e}^{\mathrm{A} / n} \mathrm{e}^{B / n}\right)^{n}\right| \leq n \times a^{(n-1) / n}\left|\mathrm{~S}{n}-\mathrm{T}{n}\right| $$ Finally, using $\mathrm{S}{n}-\mathrm{T}_{n}=-[\mathrm{A}, \mathrm{B}] / 2 n^{2}+O\left(1 / n^{3}\right)$, the product formula is verified for matrices. But the theorem also holds for self-adjoint operators.
Theorem $2.2$ (Trotter’s Theorem) If $\hat{A}$ and $\hat{B}$ are self-adjoint operators and $\hat{A}+$ $\hat{B}$ is essentially self-adjoint on the intersection $\mathscr{D}$ of their domains, then $$ \mathrm{e}^{-\mathrm{i} t(\hat{A}+\hat{B})}=s-\lim {n \rightarrow \infty}\left(\mathrm{e}^{-\mathrm{i} t \hat{A} / n} \mathrm{e}^{-\mathrm{i} t \hat{B} / n}\right)^{n} $$ If in addition $\hat{A}$ and $\hat{B}$ are bounded from below, then $$ \mathrm{e}^{-\tau(\hat{A}+\hat{B})}=s-\lim {n \rightarrow \infty}\left(\mathrm{e}^{-\tau \hat{A} / n} \mathrm{e}^{-\tau \hat{B} / n}\right)^{n} $$ The operators need not be bounded and the convergence is with respect to the strong operator topology. For operators $\hat{A}{n}$ and $\hat{A}$ on a common domain $\mathscr{D}$ in the Hilbert space, we have s- $\lim {n \rightarrow \infty} \hat{A}{n}=\hat{A}$ iff $\left|\hat{A}{n} \psi-\hat{A} \psi\right| \rightarrow 0$ for all $\psi \in \mathscr{D}$. Formula (2.27) is used in quantum mechanics, and formula $(2.28)$ finds its application in statistical physics and the Euclidean formulation of quantum mechanics [16].
Let us assume that $\hat{H}$ can be written as $\hat{H}=\hat{H}{0}+\hat{V}$ and apply the product formula to the evolution kernel in (2.22). With $\varepsilon=t / n$ and $\hbar=1$, we obtain $$ \begin{aligned} K\left(t, q^{\prime}, q\right) &=\lim {n \rightarrow \infty}\left\langle q^{\prime}\left|\left(\mathrm{e}^{-\mathrm{i} \varepsilon \hat{H}{0}} \mathrm{e}^{-\mathrm{i} \varepsilon \hat{V}}\right)^{n}\right| q\right\rangle \ &=\lim {n \rightarrow \infty} \int \mathrm{d} q_{1} \cdots \mathrm{d} q_{n-1} \prod_{j=0}^{j=n-1}\left|q_{j+1}\right| \mathrm{e}^{-\mathrm{i} \varepsilon \hat{H}{0}} \mathrm{e}^{-i \varepsilon \hat{V}}\left|q{j}\right\rangle \end{aligned} $$ where we repeatedly inserted the resolution of the identity $(2.21)$ and denoted the initial and final point by $q_{0}=q$ and $q_{n}=q^{\prime}$, respectively. The potential $\hat{V}$ is diagonal in position space such that $$ \left\langle q_{j+1}\left|\mathrm{e}^{-\mathrm{i} \varepsilon \hat{H}{0}} \mathrm{e}^{-\mathrm{i} \varepsilon \hat{V}}\right| q{j}\right\rangle=\left\langle q_{j+1}\left|\mathrm{e}^{-\mathrm{i} \varepsilon \hat{H}{0}}\right| q{j}\right\rangle \mathrm{e}^{-\mathrm{i} \varepsilon V\left(q_{j}\right)} $$
statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYS8302
物理代写|量子场论代写Quantum field theory代考|Symmetric Tensors
There is a very important twist to the notion of tensor product when one considers systems composed of several, say $n$ identical particles. Identical particles are indistinguishable from each other, even in principle. If an electron in motion scatters on an electron at rest, two moving electrons come out of the experiment and there is no way telling which of them was the electron at rest (and it can be argued that the question may not even make sense). This has to be built in the model. To this aim, we consider a Hilbert space $\mathcal{H}$ with a basis $\left(e_{i}\right){i \geq 1}$, a given integer $n$ and we denote by $\mathcal{H}{n}$ the tensor product of $n$ copies of it, in the above sense. We observe that a permutation $\sigma$ of ${1,2, \ldots, n}$ induces a transformation of $\mathcal{H}{n}$, simply by transforming the basis element $\bigotimes{k \leq n} e_{i_{k}}$ into $\bigotimes_{k \leq n} e_{i_{\sigma(k)}}$. If an element $x$ of $\mathcal{H}_{n}$ describes the state of a system consisting of $n$ identical particles, its image under this transformation describes the same particle system, so that it must be of the type $\lambda x$ for
$\lambda \in \mathbb{C}$. Since the transform of $x$ has the same norm as $x$ then $|\lambda|=1$, that is the transform of $x$ differs from $x$ only by a phase. 10
In Part I of this book we consider the simplest case where the transform of each state $x$ is $x$ itself. Particles with this property are called bosons. 11 These are not the most common and interesting particles, but must be understood first. Later on, we will meet fermions, which comprise most of the important particles (and in particular electrons). Let us denote by $\mathrm{S}_{n}$ the group of permutations of ${1, \ldots, n}$.
物理代写|量子场论代写Quantum field theory代考|Creation and Annihilation Operators
We define and study the operators $A_{n}(\xi)$ and $A_{n}^{\dagger}(\eta)$, which will play a crucial role in the next section. It is convenient here to define $\mathcal{H}{0, s}:=\mathbb{C}$. For $n \geq 1$ and $\xi \in \mathcal{H}$ the operator $A{n}(\xi): \mathcal{H}{n, s} \rightarrow \mathcal{H}{n-1, s}$ transforms an $n$-particle state into an $(n-1)$-particle state, and for this reason is called an annihilation operator. For $n \geq 0$ and $\eta \in \mathcal{H}$ the operator $A_{n}^{\dagger}(\eta)$ : $\mathcal{H}{n, s} \rightarrow \mathcal{H}{n+1, s}$ transforms an $n$-particle state into an $(n+1)$-particle state, and is called a creation operator.
In order to avoid writing formulas which are too abstract, we pick an orthonormal basis $\left(e_{i}\right){i \geq 1}$ of $\mathcal{H}$, so that an element $\eta$ of $\mathcal{H}$ identifies with a sequence $\left(\eta{i}\right){i \geq 1}{ }^{13}$ Similarly we think of an element $\alpha$ of $\mathcal{H}{n, s}$ as a symmetric tensor $\left(\alpha_{i_{1}, \ldots, i_{n}}\right){i{1}, \ldots, i_{n} \geq 1}$.
Let us then introduce an important notation: Given a sequence $i_{1}, \ldots, i_{n+1}$ of length $n+1$ we denote by $i_{1}, \ldots, \hat{i}{k}, \ldots i{n+1}$ the sequence of length $n$ where the term $i_{k}$ is omitted.
物理代写|量子场论代写Quantum field theory代考|Boson Fock Space
A relativistically correct version of Quantum Mechanics must describe systems with a variable number of particles, because the equivalence of mass and energy allows creation and destruction of particles. Let us assume that the space $\mathcal{H}$ describes a single particle. We have constructed in (3.7) the space $\mathcal{H}_{n, s}$ which describes a collection of $n$ identical particles. The boson Fock space will simply be the direct sum of these spaces (in the sense of Hilbert space) as $n \geq 0$ and will describe collections of any number of identical particles. ${ }^{16}$ We do not yet incorporate any idea from Special Relativity. The construction of the boson Fock space is almost trivial. The non-trivial structure of importance is a special family of operators described in Theorem 3.4.2.
For $n=0$ we define $\mathcal{H}{0, s}=\mathbb{C}$, and we denote by $e{\emptyset}$ its basis element (e.g. the number 1 ). The element $e_{\emptyset}$ represents the state where no particles are present, that is, the vacuum. It is of course of fundamental importance. Then we define $$ \mathcal{B}{0}=\bigoplus{n \geq 0} \mathcal{H}{n, s}, $$ the algebraic sum of the spaces $\mathcal{H}{n, s}$, where again $\mathcal{H}{n, s}$ is the space defined in (3.7). By definition of the algebraic sum, any element $\alpha$ of $\mathcal{B}{0}$ is a sequence $\alpha=(\alpha(n)){n \geq 0}$ with $\alpha(n) \in \mathcal{H}{n, s}$ and $\alpha(n)=0$ for $n$ large enough. Let us denote by $(\cdot, \cdot){n}$ the inner product on $\mathcal{H}{n, s}$. Consider $\alpha(n), \beta(n) \in \mathcal{H}{n, s}$ and $\alpha=(\alpha(n)){n \geq 0}, \beta=(\beta(n)){n \geq 0}$. We define $$ (\alpha, \beta):=\sum{n \geq 0}(\alpha(n), \beta(n)){n} . $$ The boson Fock space $\mathcal{B}$ is the space of sequences $(\alpha(n)){n \geq 0}$ such that $\alpha(n) \in \mathcal{H}{n, s}$ and $$ \left|(\alpha(n)){n \geq 0}\right|^{2}:=\sum_{n \geq 0}|\alpha(n)|^{2}<\infty, $$ where $|\alpha(n)|$ is the norm in $\mathcal{H}{n, s}$. We will hardly ever need to write down elements of $\mathcal{B}$ which are not in $\mathcal{B}{0}$.
We will somewhat abuse notation by considering each $\mathcal{H}{n, s}$, and in particular $\mathcal{H}=\mathcal{H}{1, s}$, as a subspace of $\mathcal{B}{0}$. Again, $\mathcal{H}{n, s}$ represents the $n$-particle states. Given $\xi, \eta$ in $\mathcal{H}$ we recall the operators $A_{n}(\xi)$ and $A_{n}^{\dagger}(\eta)$ of the previous section.
statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYS3101
物理代写|量子场论代写Quantum field theory代考|A First Contact with Creation and Annihilation Operators
High-energy interacting particles create other particles, and a relativistic theory must consider multiparticle systems, where the number of particles may vary.
Let us describe what is probably the simplest example of a multiparticle system. ${ }^{67}$ The “particles” are as simple as possible. ${ }^{68}$
Consider a separable Hilbert space with an orthonormal basis $\left(e_{n}\right){n \geq 0}$. The idea is that the state of the system is described by $e{n}$ when the system consists of $n$ particles. The important structure consists of the operators $a$ and $a^{\dagger}$ defined on the domain $$ \mathcal{D}=\left{\sum_{n \geq 0} \alpha_{n} e_{n} ; \sum_{n \geq 0} n\left|\alpha_{n}\right|^{2}<\infty\right} $$ by $$ a\left(e_{n}\right)=\sqrt{n} e_{n-1} ; a^{\dagger}\left(e_{n}\right)=\sqrt{n+1} e_{n+1} . $$ The definition of $a\left(e_{n}\right)$ is to be understood as $a\left(e_{0}\right)=0$ when $n=0$. The reason for the factors $\sqrt{n}$ and $\sqrt{n+1}$ is not intuitive, and will become clear only gradually. The notation is consistent, since for each $n, m$, $$ \left(e_{n}, a\left(e_{m}\right)\right)=\sqrt{m} \delta_{n}^{m-1}=\sqrt{m} \delta_{n+1}^{m}=\left(a^{\dagger}\left(e_{n}\right), e_{m}\right) $$ where $\delta_{n}^{m}$ is the Kronecker symbol (equal to 1 if $n=m$ and to 0 otherwise). Exercise 2.17.1 Prove that $a^{\dagger}$ is the adjoint of $a$. Prove in particular that if $|(y, a(x))| \leq$ $C|x|$ for $x \in \mathcal{D}$ then $y \in \mathcal{D}$.
Exercise 2.17.2 Prove that for each $\lambda \in \mathbb{C}$ the operator $a$ has an eigenvector with eigenvalue $\lambda$. Can this happen for a symmetric operator? It should be obvious from $(2.88)$ that $$ a^{\dagger} a\left(e_{n}\right)=n e_{n} ; a a^{\dagger}\left(e_{n}\right)=(n+1) e_{n} . $$ Let us then consider the self-adjoint operator ${ }^{69}$ $$ N:=a^{\dagger} a . $$ Thus $N\left(e_{n}\right)=n e_{n}$. Since a system in state $e_{n}$ has $n$ particles, the observable corresponding to this operator is “the number of particles”. The operator $N$ is therefore called the number operator. As another consequence of $(2.90)$, $$ \left[a, a^{\dagger}\right]\left(e_{n}\right)=e_{n t} $$
物理代写|量子场论代写Quantum field theory代考|The Harmonic Oscillator
The fundamental structure outlined in the previous section is connected to an equally fundamental system, the harmonic oscillator. A classical one-dimensional harmonic oscillator of angular frequency ${ }^{73} \omega$ consists of a point of mass $m$ on the real line which is pulled back to the origin with a force $m \omega^{2}$ times the distance to the origin. The quantum version of this system is the space $\mathcal{H}=L^{2}(\mathbb{R})$ with Hamiltonian $$ H:=\frac{1}{2 m}\left(P^{2}+\omega^{2} m^{2} X^{2}\right), $$ where $P$ and $X$ are respectively the momentum and the position operators of Section $2.5$. This is the Hamiltonian $(2.79)$ in the case where $V(x)=m \omega^{2} x^{2} / 2$. That this formula provides a quantized version of the classical harmonic oscillator is not obvious at all. ${ }^{74}$ We will explain in Section $6.6$ the systematic procedure of “canonical quantization” to discover formulas such as (2.77) or (2.95). This procedure is by no means a proof of anything, and the resulting formulas are justified only by the fact that they provide a fruitful model. So there is little harm to accept at this stage that the formula (2.95) is indeed fundamental. We have not proved yet that this formula defines a self-adjoint operator, but this is a consequence of the analysis below.
Exercise 2.18.1 Prove that a symmetric operator which admits an orthonormal basis of eigenvectors is self-adjoint. Hint: If we denote by $\left(e_{n}\right)$ an orthonormal basis of eigenvectors, and $\lambda_{n}$ the eigenvalue of $e_{n}$, the natural domain $\mathcal{D}$ of the operator is $$ \left{x=\sum_{n} x_{n} e_{n} ; \sum_{n}\left(1+\left|\lambda_{n}\right|^{2}\right)\left|x_{n}\right|^{2}<\infty\right} $$ The program for this section is first to find a basis of eigenvectors for the Hamiltonian (2.95), and then to examine how some classical quantities transform under quantization.
物理代写|量子场论代写Quantum field theory代考|Tensor Products
The present section is standard material, but our presentation attempts to balance rigor and readability.
Principle 6 If the states of two systems $\mathcal{S}{1}$ and $\mathcal{S}{2}$ are represented by the unitary rays in two Hilbert spaces $\mathcal{H}{1}$ and $\mathcal{H}{2}$ respectively, the appropriate Hilbert space to represent the system consisting of the union of $\mathcal{S}{1}$ and $\mathcal{S}{2}$ is the tensor product $\mathcal{H}{1} \otimes \mathcal{H}{2}$.
Our first task is to describe this space. ${ }^{2}$ A mathematician would love to see an “intrinsic” definition of this tensor product, a definition that does not use bases or a special representation of these Hilbert spaces. This can be done elegantly as in e.g. Dimock’s book [23]. We shall not enjoy this piece of abstraction and we shall go the ugly way.
If $\left(e_{n}\right){n \geq 1}$ is an orthonormal basis of $\mathcal{H}{1}$ and $\left(f_{n}\right){n \geq 1}$ is an orthonormal basis of $\mathcal{H}{2}$ then the vectors $e_{n} \otimes f_{m}$ constitute an orthonormal basis of $\mathcal{H}{1} \otimes \mathcal{H}{2}$, which is thus the set of vectors of the type $\sum_{n, m \geq 1} a_{n, m} e_{n} \otimes f_{m}$ where the complex numbers $a_{n, m}$ satisfy $\sum_{n, m \geq 1}\left|a_{n, m}\right|^{2}<\infty$. Here the quantity $e_{n} \otimes f_{m}$ is just a notation, which is motivated by the fact that for $x=\sum_{n \geq 1} \alpha_{n} e_{n} \in \mathcal{H}{1}$ and $y=\sum{n \geq 1} \beta_{n} f_{n} \in \mathcal{H}{2}$ one defines $x \otimes y \in$ $\mathcal{H}{1} \otimes \mathcal{H}{2}$ by $$ x \otimes y=\sum{m, n \geq 1} \alpha_{n} \beta_{m} e_{n} \otimes f_{m} . $$ When either $\mathcal{H}{1}$ or $\mathcal{H}{2}$, or both, are finite-dimensional, the definition is modified in the obvious manner.
Exercise 3.1.1 When $\mathcal{H}{1}$ and $\mathcal{H}{2}$ are finite-dimensional, what is the dimension of $\mathcal{H}{1} \otimes \mathcal{H}{2}$ ? How does it compare with the dimension of the usual product $\mathcal{H}{1} \times \mathcal{H}{2}$ ? When either $\mathcal{H}{1}$ or $\mathcal{H}{2}$ is infinite-dimensional, $\mathcal{H}{1} \otimes \mathcal{H}{2}$ is an infinite-dimensional Hilbert space. The important structure is the bilinear form from $\mathcal{H}{1} \times \mathcal{H}{2}$ into this space given by (3.1).
Recalling that $(x, y)$ denotes the inner product in a Hilbert space we observe the formula $$ \left(x \otimes y, x^{\prime} \otimes y^{\prime}\right)=\left(x, x^{\prime}\right)\left(y, y^{\prime}\right), $$ which is a straightforward consequence of the fact that the basis $e_{n} \otimes f_{m}$ is orthonormal. $^{3}$ The problem with our definition of the tensor product is that one is supposed to check that “it does not depend on the choice of the orthonormal basis”, a tedious task that joins similar tasks under the carpet. ${ }^{4}$ The good news is that all the identifications one may wish for are true. If both $\mathcal{H}{1}$ and $\mathcal{H}{2}$ are the space of square-integrable functions on $\mathbb{R}^{3}$, then $\mathcal{H}{1} \otimes \mathcal{H}{2}$ is the space of square-integrable functions on $\mathbb{R}^{6} .5$ This fits very well with the Dirac formalism: If $|x\rangle$ denotes the Dirac function at $\boldsymbol{x}$, (so that these generalized vectors provide a generalized basis of $\mathcal{H}{1}$, and similarly for $|\boldsymbol{y}\rangle$, then $|\boldsymbol{x}\rangle|\boldsymbol{y}\rangle$ denotes the Dirac function at the point $(\boldsymbol{x}, \boldsymbol{y}) \in \mathbb{R}^{6}$, and these generalized vectors provide a generalized basis of $\mathcal{H}{1} \otimes \mathcal{H}{2}$. Furthermore, if $f \in \mathcal{H}{1}$ and $g \in \mathcal{H}_{2}$ then $f \otimes g$ identifies with the function $(\boldsymbol{x}, \boldsymbol{y}) \mapsto$ $f(x) g(y)$ on $\mathbb{R}^{6}$.
statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYS4040
物理代写|量子场论代写Quantum field theory代考|One-parameter Unitary Groups and Stone’s Theorem
A (strongly continuous) one-parameter unitary group is simply a (strongly continuous) unitary representation of $\mathbb{R}$, that is a map which associates to $t \in \mathbb{R}$ a unitary operator $U(t)$ on Hilbert space $\mathcal{H}$ in such a manner that $$ U(s) U(t)=U(s+t), $$ and (the continuity condition) $$ \forall x, y \in \mathcal{H}, \lim _{t \rightarrow 0}(x, U(t)(y))=(x, y) . $$
The archetypical example is the operator $U(t)$ on $L^{2}(\mathbb{R})$ given for $f \in L^{2}$ and $w \in \mathbb{R}$ by $$ U(t)(f)(w)=\exp (\mathrm{i} t w / h) f(w) . $$ This is simply the operator “multiplication by the function exp(it $\cdot / \hbar) . “$ Another example is the operator $V(t)$ on $L^{2}(\mathbb{R})$ given for $f \in L^{2}$ and $w \in \mathbb{R}$ by $$ V(t)(f)(w)=f(w+t) . $$ In both cases it is a nice exercise of elementary analysis to prove that these operators are strongly continuous. These one-parameter groups are closely related by the Fourier transform. Indeed, $$ \widehat{V(t)(f)}=U(t) \hat{f} . $$ Exercise 2.14.1 Make sure you understand every detail of the proof of the important formula (2.55)
Theorem 2.14.2 (Stone’s theorem) There is a one-to-one correspondence between the strongly continuous one-parameter unitary groups on a Hilbert space $\mathcal{H}$ and the self-adjoint operators on $\mathcal{H}$. Given the unitary group $U$, the corresponding self-adjoint operator $A$ is called the infinitesimal generator of U. It is defined by the formula ${ }^{58}$ $$ A(x)=\lim _{t \rightarrow 0} \frac{h}{i}(U(t)(x)-x) . $$ and its domain $\mathcal{D}=\mathcal{D}(A)$ is the set of $x$ for which the previous limit exists.
物理代写|量子场论代写Quantum field theory代考|Time-evolution
Consider a physical system, the state of which is described by a vector in $\mathcal{H}$. Principle 5 If the system does not change with time (not in the sense that it does not evolve, but in the sense that it is not subjected to variable external influences), its evolution between time $t_{0}$ and time $t_{1}$ is described by a unitary operator ${ }^{62} U\left(t_{1}, t_{0}\right)$.
This operator depends only on $t_{1}-t_{0}$, reflecting the fact that the laws of physics are believed not to change with time ${ }^{63}$
Please observe the notation: the evolution $U\left(t_{1}, t_{0}\right)$ is from $t_{0}$ to $t_{1}$. The reason for this notation is that the evolution of the system from $t_{0}$ to $t_{1}$ and then from $t_{1}$ to $t_{2}$ is represented by $U\left(t_{2}, t_{1}\right) U\left(t_{1}, t_{0}\right)$, which also represents the same evolution as $U\left(t_{2}, t_{0}\right)$ so that as in (2.45) these two operators should differ only by a phase, $U\left(t_{2}, t_{1}\right) U\left(t_{1}, t_{0}\right)=c U\left(t_{2}, t_{0}\right)$ for some $c$ of modulus 1. Thus $U\left(t_{2}-t_{1}, 0\right) U\left(t_{1}-t_{0}, 0\right)=c U\left(t_{2}-t_{0}, 0\right)$. This means that $U(t):=$ $U(t, 0)$ should be a projective representation of $\mathbb{R}$ in $\mathcal{H}$, and on physical grounds it should be continuous in some sense. As shown in Section $2.13$ this projective representation arises from a true representation, so we can as well assume that it already is a true representation, and Stone’s theorem describes these. Therefore, there exists a self-adjoint operator $H$ on $\mathcal{H}$ such that $$ U(t)=\exp (-\mathrm{i} t H / \hbar) . $$ Probably it is worth making explicit the following fundamental point: The time-evolution of a quantum system is entirely deterministic. The minus sign in (2.75) is conventional. The reason for this convention will appear in Section 9.7. Since $h$ has the dimension of an energy times a time, $H$ has the dimension of an energy. It is called the Hamiltonian of the system. Although this is certainly not obvious ${ }^{64}$ The Hamiltonian should be thought of as representing the energy of the system. A consequence of (2.75) is that if $\psi$ belongs to the domain of $H$, then, by the second part of $(2.64), \psi(t):=U(t)(\psi)$ satisfies the equation $$ \psi^{\prime}(t)=-\frac{\mathrm{i}}{h} H(\psi(t)) \text {. } $$
物理代写|量子场论代写Quantum field theory代考|Schrödinger and Heisenberg Pictures
The previous description of an evolving state $\psi(t)$ and of time-independent operators is called the Schrödinger picture.
We have seen that we may be able to improve matters by re-shuffling the state space using a unitary transformation. A fundamental idea is to try this using a time-dependent unitary transformation $V(t)$, replacing the state $\psi$ by $V(t)(\psi)$, and replacing the operator $A$ by $A(t)=V(t) A V(t)^{-1}$. Here we use a first simple implementation, with $V(t)=$ $\exp ($ it $H / \hbar)=U(t)^{-1}$. This is called the Heisenberg picture. In the Heisenberg picture the state $\psi(t)$ is replaced by $V(t)(\psi(t))=U(t)^{-1} U(t)(\psi)=\psi$, so that states do not change with time. On the other hand, an operator $A$ is replaced by the operator $$ A(t)=U(t)^{-1} A U(t)=\exp (\mathrm{i} t H / \hbar) A \exp (-\mathrm{i} t H / \hbar) . $$ Suppose that at time $t=0$ the system is in state $\psi$. Then, in the Schrödinger picture, the average value of $A$ at time $t$ is given by $$ (\psi(t), A \psi(t))=(U(t) \psi, A U(t) \psi)=\left(\psi, U(t)^{-1} A U(t) \psi\right)=(\psi, A(t) \psi), $$ where the last expression is the same quantity in the Heisenberg picture. Thus these two pictures are fortunately consistent with each other.
We may wonder why there is anything to gain by moving from the Schrödinger picture, where the simpler objects, the states, evolve, but where the complicated objects, the operators, are constant, to the Heisenberg picture where the simpler objects are constant, but the complicated ones evolve. One reason is that while it is correct in principle that the states, which are simply vectors of $\mathcal{H}$ are simpler objects than operators, it will often happen that the operators of interest have a simple description, while the states have a very complicated one. Another reason will be given at the end of the present section.
statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。
我们提供的量子场论Quantum field theory及其相关学科的代写，服务范围广, 其中包括但不限于:
Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(Generalized) Linear Models 广义线性模型
Statistical Machine Learning 统计机器学习
Longitudinal Data Analysis 纵向数据分析
Foundations of Data Science 数据科学基础
物理代写|量子场论代写Quantum field theory代考|PHYS7076
物理代写|量子场论代写Quantum field theory代考|Projective versus True Unitary Representations
Let us start the discussion of the concepts involved in Definitions $2.10 .1$ and 2.10.3. The word “unitary” refers of course to the fact that each of the operators $U(a)$ is unitary. Unless mentioned otherwise, all representations are unitary, so that we shall nearly always omit the word “unitary”, and the expressions “representation” and “projective representations” have to be understood by default as “unitary representation” and “projective unitary representations”.
To insist that a representation satisfies $r(a, b)=1$ for all $a, b$ we will sometimes say true representation, even though throughout the book, the word “representation” means “true representation”. When we consider a representation that is only a projective representation we will always say so explicitly. It is most important to understand the relationship between representations and projective representations.
The concept of “representation” is far more restrictive than the concept of “projective representation”.
From the point of view of mathematics, the nice objects are representations. The study of group representations is a vast subject in mathematics.
From the point of view of Quantum Mechanics, the natural objects are projective representations.
The following explains an important relationship between representations and projective representations.
Definition 2.11.1 Given a true representation $V$ of $G$, and for $a \in G$ a number $\lambda(a)$ of modulus 1 , the formula $$ U(a):=\lambda(a) V(a) $$ defines a projective representation, since (2.45) holds for the function $$ r(a, b)=\lambda(a b) /(\lambda(a) \lambda(b)) . $$ When this is the case we will say that the projective representation Uarises from the true representation $V$.
More generally, there is an important idea behind this definition: two projective representations $U, U^{\prime}$ such for each $a \in G$ one has $U(a)=\lambda(a) U^{\prime}(a)$ for some complex number $\lambda(a)$ with $|\lambda(a)|=1$ are to be thought of as “the same projective representation”.
物理代写|量子场论代写Quantum field theory代考|Mathematicians Look at Projective Representations
This material is not needed to follow the main story. It assumes that you know some very basic group theory. A map $U$ from $G$ to the group $\mathcal{U}(\mathcal{H})$ of unitary transformations of $\mathcal{H}$ is a true representation if and only if it is a group homomorphism. The group $\mathcal{U}(\mathcal{H}$ ) has a remarkable subgroup, the subgroup consisting of the transformations $\lambda 1$ with $|\lambda|=1$. Let us denote by $\mathcal{U}{p}(\mathcal{H})$ the quotient of $\mathcal{U}(\mathcal{H})$ by this subgroup, and by $\Phi$ the quotient map $\mathcal{U}(\mathcal{H}) \rightarrow \mathcal{U}{p}(\mathcal{H})$. Thus the elements of $\mathcal{U}{p}(\mathcal{H})$ are unitary operators “up to a phase”, i.e. up to a multiplicative constant of modulus 1 . It is immediate to check that a map $U$ from a group $G$ into $\mathcal{U}(\mathcal{H})$ is a projective representation in the sense of Definition $2.10 .1$ if and only $\Phi \circ U$ is a group homomorphism from $G$ to $\mathcal{U}{p}(\mathcal{H})$. The important object is thus the map $\Phi \circ U$. Accordingly, mathematicians define a projective representation as a group homomorphism from $G$ to $\mathcal{U}{p}(\mathcal{H})$. This formalizes the idea that two projective representations $U$ and $U^{\prime}$ such that $U(a)=\lambda(a) U^{\prime}(a)$ “are the same projective representation” (because this is the case if and only if $\Phi \circ U=\Phi \circ U^{\prime}$ ). Another benefit of this approach is that it becomes natural to define “continuous projective representations”, a topic which is investigated in Section A.2. In mathematical language, the fundamental question, is, given a projective representation $U$ of $G$, that is a group homomorphism from $G$ to $\mathcal{U}{p}(\mathcal{H})$, whether there exists a true representation $V$, that is a group homomorphism from $G$ to $\mathcal{U}(\mathcal{H})$, such that $U=\Phi \circ V$.
物理代写|量子场论代写Quantum field theory代考|Projective Representations of R
We do not investigate in detail how true and projective representations are related in general, but we examine this question in the centrally important case $G=\mathbb{R}$. However, we must first discuss a technical question. In the cases of greatest interest, $G$ is a topological group, and to avoid pathologies, one requires also a mild continuity assumption.
Definition 2.13.1 The map $a \mapsto U(a)$ which associates to each element $a$ of $G$ a unitary operator $U(a)$ is called strongly continuous if for each $x \in \mathcal{H}$ the map $a \mapsto U(a)(x)$ from $G$ to $\mathcal{H}$ is continuous.
The topology on $\mathcal{H}$ is the topology induced by its norm, so the condition of strong continuity means that for each $x \in \mathcal{H}$ the norm $\left|U(a)(x)-U\left(a_{0}\right)(x)\right|$ goes to 0 as $a \rightarrow a_{0}$. Despite the adjective “strong”, this condition is much weaker than the continuity of the map $a \mapsto U(a)$ in the operator norm.
A simple but instructive example of a representation is the case where $G=\mathbb{R}$, $\mathcal{H}=L^{2}(\mathbb{R})$ and $U(a)(f) \in L^{2}(\mathbb{R})$ is the function $w \mapsto f(w-a)$. The map $a \mapsto U(a)$ is not continuous when the space of unitary operators is provided with the topology induced by the operator norm but it is strongly continuous (as one sees by approximating $f$ with a continuous function of bounded support).
When the map $a \mapsto U(a)$ is strongly continuous, then for $x, y \in \mathcal{H}$ the map $a \mapsto(x, U(a)(y))$ is continuous. This apparently weaker condition is equivalent to strong continuity. To prove this, assume the weaker condition. Then as $a \rightarrow a_{0},\left(U(a)(x), U\left(a_{0}\right)(x)\right)$ tends to the square of the norm of $U\left(a_{0}\right)(x)$, and since both vectors $U(a)(x)$ and $U\left(a_{0}\right)(x)$ have the same norm they become close to each other (as follows from the relation $\left.|u-v|^{2}=|u|^{2}+|v|^{2}-2 \operatorname{Re}(u, v)\right)$
